Let u, v, w be non-coplanar vectors and p, q are real numbers, then the equality [3u pv pw] - [pv w qu] - [2w qv qu] = 0 holds for: - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

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Let u, v, w be non-coplanar vectors and p, q are real numbers, then the equality [3 u p v p w] - [p v w q u] - [2 w q v q u] = 0 holds for:

Solution & Step-by-step Explanation

Using properties of scalar triple product: 3 p^{2} [u v w] - pq [v w u] - 2 q^{2} [w v u] = 0 .Since [u v w] = [v w u] = - [w v u], the equation becomes: (3 p^{2} - pq + 2 q^{2}) [u v w] = 0 .As vectors are non-coplanar, [u v w] \neq = 0 . ⟹ 3 p^{2} - pq + 2 q^{2} = 0 .This is a quadratic form. For p, q \in R, this is only zero if p = 0, q = 0 .Thus, exactly one value of (p, q) .

Try it yourself before checking the explanation above.

Let u, v, w be non-coplanar vectors and p, q are real numbers, then the equality [3 u p v p w] - [p v w q u] - [2 w q v q u] = 0 holds for:

exactly one value of (p, q)

exactly two values of (p, q)

more than two but not all values of (p, q)

all values of (p, q)

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